Electronic – How to we neglect copper losses in open circuit test then neglect core losses in short circuit test of transformer

'Copper loss' is I²R power loss in the windings due to current flowing through them. As this current increases with higher loading, so copper loss also increases as loading increases.

Eddy current loss is power loss in the magnetic core due to current induced into it (each lamination in the core is effectively a shorted turn, but the silicon steel has relatively high resistance which keeps the current down). The induced voltage - and thus current and power loss - doesn't change with loading because the magnetic flux in the core doesn't change. This is also I²R loss, but it is not copper loss.

To summarize, there are two places in the transformer where I²R losses occur - in the magnetic core and in the windings. However, only the windings have a power loss proportional to load current. That loss can rightly be called 'copper loss' because only the windings are made of copper.

You're on the right track, but we have to do a little more calculation. The short circuit test data allows us to calculate the resistance. In the schematic below, we apply some voltage \$V_1\$ and measure the current \$I_1\$ on the primary side. What we can then calculate is \$Z_1\$, the equivalent series impedance which is \$R_1 + j X_1\$.

^{simulate this circuit – Schematic created using CircuitLab}

In particular, we have $$R_1 = \frac{P_1}{|I_1|^2}$$ In this case, we have \$R_1=37947/225^2 = 0.75\Omega\$ where \$R_1\$ is the equivalent series resistance on the primary side. If we want to be pedantic, although it's not needed for this problem, we could calculate \$X_1\$ because we know that $$|Z_1|^2 = \left(\frac{|V_1|}{|I_1|}\right)^2 = R_1^2 + X_1^2$$ To translate that to a secondary side equivalent series resistance, we multiply by the square of the turns ratio: \$R_2=R_1\left(\frac{50000V}{10000V}\right)^2=18.739\Omega\$.

^{simulate this circuit}

There are a number of equivalent ways to express efficiency, but in this case the most useful form is $$\eta=\frac{P_{out}}{P_{out}+W_C+W_I}$$ where \$W_C\$ is the copper loss and \$W_I\$ are hysteresis and eddy losses, which are usually combined and called "iron losses" or "core losses". The other key thing to remember is that iron losses are not dependent on load current while copper losses are: \$W_C = |I|^2R\$ Substituting that into the equation gives $$\eta=\frac{P_{out}}{P_{out}+|I_2|^2R_2+W_I}$$ A little algebra lets us solve for \$W_I\$. $$W_I=\frac{P_{out}}{\eta}-P_{out}-|I_2|^2R_2$$ Since we're given efficiency at unity power factor at nominal we can simply plug in the numbers and calculate the result: $$W_I = \frac{2500 kW}{0.9766} - 2500kW - (50A)^2(18.739\Omega)$$ Here, the 50A is the secondary current at nominal (2500VA/50kV). We could equivalently used the primary current at nominal (250A = 2500VA/10kV) and \$R_1\$. Mathematically, they're identical. I calculate \$W_I = 13.054kW\$ which is, directly, the open circuit power.

One can also apply this the other way around. That is, if we are given the open circuit power, we have \$W_I\$ and can, given the efficiency \$\eta\$, calculate \$W_C\$. Since we know that this is at the rated current, which is also how the short circuit test is conducted (if you can forgive the pun) we can extract \$R2\$.